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It is natural to substitute $a=2^x$. Then the following code

var('a');
# a will be 2^x at some point 
A = matrix( 4, 4, [ 0,1,1,1, 1,0,a,a, 1,a,0,a, 1,a,a,0 ] )
print "The matrix A is\n%s" % A
print "det(A) = %s" % A.det()
print "The inverse of A is\n%s" % A.inverse()
print "The sum of the entries in A.inverse() is %s" % sum( A.inverse().list() )

delivers:

a
The matrix A is
[0 1 1 1]
[1 0 a a]
[1 a 0 a]
[1 a a 0]
det(A) = -3*a^2
The inverse of A is
[-2/3*a    1/3    1/3    1/3]
[   1/3 -2/3/a  1/3/a  1/3/a]
[   1/3  1/3/a -2/3/a  1/3/a]
[   1/3  1/3/a  1/3/a -2/3/a]
The sum of the entries in A.inverse() is -2/3*a + 2

So for a value of $a$ different from zero we have $\det A\ne 0$. The vanishing ot the sum of the entries of $A^{-1}$ is then what we want, this happens for $a=3$, so the corresponding value of $x$ is the solution of $2^x=a=3$, this is $$x=\log_23\ .$$

P.S.

sage: A.inverse().subs( {a:3} )
[  -2  1/3  1/3  1/3]
[ 1/3 -2/9  1/9  1/9]
[ 1/3  1/9 -2/9  1/9]
[ 1/3  1/9  1/9 -2/9]
sage: sum(_.list())
0